AngouriMath
SolveAnExponentialOfAnInverseHyperbolicTangentAlgebraically(AngouriMath.Entity,AngouriMath.Entity.Variable)
Method (no overloads)
Summary
An exponential of a whole multiple of an inverse hyperbolic tangent of a linear, written
algebraically:e^(n artanh(u)) arrives as e^((n/2) ln((1 + u)/(1 - u))) , and
is(1 + u)^n (1 - u^2)^(-n/2) for |u| < 1 , where the inverse is real. The
integrand so written is asked as a question of its own.
algebraically:
is
integrand so written is asked as a question of its own.
Remarks
The rules after it timed out on Rubi's 7.3.6 e^(artanh(a x)) x^2/(c - a^2 c x^2)^2 ,
which written so is(1 + a x) x^2/(c^2 (1 - a^2 x^2)^(5/2)) , answered in half a
second, and one^(4 artanh(a x))/(c - a^2 c x^2)^2 , a rational function.
https://github.com/asc-community/AngouriMath/issues/718
which written so is
second, and on
https://github.com/asc-community/AngouriMath/issues/718
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